# NCERT Solutions Class 12 Maths (Three Dimensional Geometry) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-11 (Three Dimensional Geometry) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

Answer. Let OA be the line joining the origin, O (0, 0, 0) and the point A(2, 1, 1) Also, let BC be the line joining the points, B (3, 5, -1) and C (4, 3, -1). The direction ratios of OA are 2, 1, and 1 and of BC are (4 - 3) = 1, (3 - 5) = -2 and (-1 + 1) = 0 OA is perpendicular to BC, if ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0$ $\therefore {a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=2×1+1\left(-2\right)+1×0=2-2=0$ Thus, OA is perpendicular to BC.

Q2. If  are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are ${m}_{1}{n}_{2}-{m}_{2}{n}_{1},{n}_{1}{l}_{2}-{n}_{2}{l}_{1},{l}_{1}{m}_{2}-{l}_{2}{m}_{1}$.

Answer. It is given that  are the direction cosines of two mutually perpendicular lines. Therefore, $\begin{array}{l}{l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}=0\\ {l}_{1}^{2}+{m}_{1}^{2}+{n}_{1}^{2}=1\\ {l}_{2}^{2}+{m}_{2}^{2}+{n}_{2}^{2}=1\end{array}$ Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines $\begin{array}{l}\therefore ‖{}_{1}+m{m}_{1}+m{n}_{1}=0\\ ‖{}_{2}+m{m}_{2}+{m}_{2}=0\end{array}$ $\therefore \frac{l}{{m}_{1}{n}_{2}-{m}_{2}{n}_{1}}=\frac{m}{{n}_{1}{l}_{2}-{n}_{2}{I}_{1}}=\frac{n}{{l}_{1}{m}_{2}-{l}_{2}{m}_{l}}$ $⇒\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{1}{l}_{2}-{n}_{2}{I}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{i}\right)}^{2}}$ $⇒\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{1}{l}_{2}-{n}_{2}{I}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{2}\right)}^{2}}$ $=\frac{{t}^{2}+{m}^{2}+{n}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{i}\right)}^{2}}$ l, m, n are the direction cosines of the line. $\therefore {l}^{2}+{m}^{2}+{n}^{2}=1\dots \left(5\right)$ It is known that, $\left({l}_{1}^{2}+{m}_{1}^{2}+{n}_{1}^{2}\right)\left({l}_{2}^{2}+{m}_{2}^{2}+{n}_{2}^{2}\right)-{\left({l}_{1}{l}_{2}+{m}_{1}{m}_{2}+{n}_{1}{n}_{2}\right)}^{2}$ $={\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{1}\right)}^{2}$ From (1), (2), and (3), we obtain $⇒1.1-0={\left({m}_{1}{n}_{2}+{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{l}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{1}\right)}^{2}$ ${\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}+{\left({n}_{1}{l}_{2}-{n}_{2}{I}_{1}\right)}^{2}+{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{1}\right)}^{2}=1$ Substituting the values from equations (5) and (6) in equation (4), we obtain $\begin{array}{l}\frac{{l}^{2}}{{\left({m}_{1}{n}_{2}-{m}_{2}{n}_{1}\right)}^{2}}=\frac{{m}^{2}}{{\left({n}_{2}{I}_{2}-{n}_{2}{l}_{1}\right)}^{2}}=\frac{{n}^{2}}{{\left({l}_{1}{m}_{2}-{l}_{2}{m}_{1}\right)}^{2}}=1\\ ⇒l={m}_{1}{n}_{2}-{m}_{2}{n}_{1},m={n}_{1}{l}_{2}-{n}_{2}{I}_{11},n={l}_{1}{m}_{2}-{l}_{2}{m}_{1}\end{array}$ Thus, the direction cosines of the required line are .

Q3. Find the angle between the lines whose direction ratios a, b, c and $b-c,c-a,a-b$.

Answer. The angle Q between the lines with direction cosines a, b, c and b-c, c-a, a-b, is given by, $\begin{array}{l}\mathrm{cos}Q=|\frac{a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}+\sqrt{\left(b-c{\right)}^{2}+\left(c-a{\right)}^{2}+\left(a-b{\right)}^{2}}}|\\ ⇒Q={\mathrm{cos}}^{-1}0\\ ⇒Q={90}^{\circ }\end{array}$ Thus, the angle between the lines is ${90}^{\circ }$

Q4. Find the equation of a line parallel to x-axis and passing through the origin.

Answer. The line parallel to x-axis and passing through the origin is x-axis itself. Let A be a point on x-axis . Therefore, the coordinates of A are given by (a, 0, 0) where $a\in R$ Direction ratios of OA are (a - 0) = a, 0, 0 The equation of OA is given by, $\begin{array}{l}\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\\ ⇒\frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\end{array}$ Thus, the equation of line parallel to x-axis and passing through origin is

Q5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer. The coordinates of A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively. The direction ratios of AB are (4, -1) = 3, (5 - 2)=3, and (7 - 3) = 4 The direction ratios of CD are  It can be seen that, $\frac{{a}_{1}}{{a}_{2}}=\frac{{b}_{1}}{{b}_{2}}=\frac{{c}_{1}}{{c}_{2}}=\frac{1}{2}$ Therefore, AB is parallel to CD. Thus, the angle between AB and CD is either .

Q6. If the lines  are perpendicular, find the value of k.

Answer. The direction of ratios of the lines,  are -3, 2k, 2 and 3k, 1, -5 respectively. It is known that two lines with direction ratios,  are perpendicular, if ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=$ = 0. $\begin{array}{l}\therefore -3\left(3k\right)+2k×1+2\left(-5\right)=0\\ ⇒-9k+2k-10=0\\ ⇒7k=-10\\ ⇒k=\frac{-10}{7}\end{array}$ Therefore, for $k=-\frac{10}{7}$, the given lines are perpendicular to each other.

Q7. Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\stackrel{\to }{r}\cdot \left(\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}\right)+9=0$.

Answer. The position vector of the point (1, 2, 3) is ${\stackrel{\to }{r}}_{1}=\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}$ The direction ratios of the normal to the plane, $\stackrel{\to }{r}\cdot \left(\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}\right)+9=0$ are 1, 2, and -5 and the normal vector is $\overline{N}=\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}$. The equation of a line passing through a point and perpendicular to the given plane is given by, $\stackrel{\to }{l}=\stackrel{\to }{r}+\lambda \stackrel{\to }{N},\lambda \in R$

Q8. Find the equation of the plane passing through (a, b, c) and parallel to the plane $\stackrel{\to }{r}\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2$.

Answer. Any plane parallel to the plane, ${\stackrel{\to }{r}}_{1}\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2$ is of form $\stackrel{\to }{r}\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=\lambda$ The plane passes through the point (a, b, c). Therefore, the position vector of this point is $\stackrel{\to }{r}=a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}$ Therefore, equation (1) becomes $\begin{array}{l}\left(a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}\right)\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=\lambda \\ ⇒a+b+c=\lambda \end{array}$ Substituting $\lambda =a+b+c$ in equation (1), we obtain $\stackrel{\to }{r}\cdot \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=a+b+c$ This is the vector equation of the required plane. Substituting $\stackrel{\to }{r}=x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}$ in equation (2), we obtain

Q9. Find the shortest distance between lines